Prove or disprove that any three members of a family of parallelograms intersect Given a family of parallelograms such that the corresponding edges of all members are parallel and any two members of this family intersect. Can we conclude that any three members of this family intersect? 
Intuitively I thought the answer should be positive. But I am having difficulty to write a rigorous proof (or find a counter example). Can anyone provide help me with this? Thanks
 A: Hint:


*

*We will say that a family of parallelograms has property $P$ if any two elements intersecting implies that any three elements intersect. We want to prove that all the families of parallelograms have this property.

*Use a shear mapping (a kind of linear transformation) to make all the parallelograms into rectangles. Prove that a family of parallelograms has property $P$ if and only if its transformed instance has property $P$.

*To prove that this is true for families of rectangles, introduce a coordinate system such that the sides of rectangles are parallel to the axes. Observe that two rectangles intersect if and only if both their orthogonal projections onto axes intersect.

*Note that because the sides are parallel to the axes the two dimensions are independent. Solve the problem for a single dimension (i.e. segments on a line), and combine twice to get result for the rectangles.

*In fact we could have skipped the transformation step, but then you would have to take non-standard coordinate system and non-orthogonal projections (i.e. projections along the vector parallel to the sides). Pick the solution which is more convenient for you.


I hope this helps $\ddot\smile$
